Solve the Problem $R$ Of a Projectile Is Related to the Initial Velocity

Solve the problem.
-The range $R$ of a projectile is related to the initial velocity $v$ and projection angle $\theta$ by the equation $\mathrm { R } = \frac { \mathrm { v } ^ { 2 } \sin 2 \theta } { \mathrm { g } }$ , where $\mathrm { g }$ is a constant. How is $\mathrm { dR } / \mathrm { d }$ t related to $\mathrm { dv } / \mathrm { dt }$ and $\mathrm { d } \theta / \mathrm { dt }$ if neither $v$ nor $\theta$ is constant?

A) $\frac { d R } { d t } = \frac { v } { g } \left( v \cos 2 \theta \frac { d \theta } { d t } + 2 \sin 2 \theta \frac { d v } { d t } \right)$
B) $\frac { d R } { d t } = \frac { 1 } { g } \left( v \cos 2 \theta \frac { d v } { d t } + \sin 2 \theta \frac { d \theta } { d t } \right)$
C) $\frac { d R } { d t } = \frac { 2 v } { g } \left( v \cos 2 \theta \frac { d \theta } { d t } + \sin 2 \theta \frac { d v } { d t } \right)$
D) $\frac { \mathrm { dR } } { \mathrm { dt } } = \frac { 1 } { \mathrm {~g} } \left( 4 \mathrm { v } \cos 2 \theta \frac { \mathrm { d } \theta } { \mathrm { dt } } \frac { \mathrm { dv } } { \mathrm { dt } } \right)$